The effect of the skin-zone on non-Newtonian oil fields

Calculation of production of non-Newtonian oil wells

In non-Newtonian oil fields, we can use the following reporting methodology to find and predict the productivity of wells.

We know that the production of wells in operation in anomalous fluid oil fields is calculated using the following formula:

2Trkh[(Pk-Pq)-i0(Rk-Rq)]

*

Here the thickness of h-layer ,k-layer conductivity ,P k - contour pressure , P q -wellbore pressure , R k - the radius of the zone around the well ,R q - the radius of the well ,S-skin factor , ^-dynamic viscosity , i₀ = ^=- initial pressure gradient, t ₀ -initial touch voltage

It is clear from the equation that R k >>R q . Therefore, it can be ignored in the report:

2nkh [(Pk - Pq) - ioRk] " MH

Now let's run the reports based on the given values:

h=10m, k=0.1*10 -12 m ² , P k =120*10 5 N/m ² , P q =80*10 5 N/m ² , R k =1000 m , i 0 =220 N/m ² ,μ=15*10 -3 N*s/m ²

Based on the given prices, let's calculate the volume production:

• 1)    When S=1:

_ 2nkh[(P_k - P _q ) - i₀R_k^ _ 2 • 3,14 • 0,1 • 10^-12 • 10[(120 - 80) • 10⁵ - 220 • 1000]

• ^Q       ^[/nR ^ + s]                      15 • 10^-3 [Zn^^^ + 1]

= 0.00083 m 3 /s = 71.712 m 3 /day

• 2)    When S=5:

_ 2тткй[(Р _к - P _q ) - i ₀ R _k] _ 2 • 3,14 • 0,1 • 10^-12 • 10[(120 - 80) • 10⁵ - 220 • 1000]

• ^ =           R_k J =                   _1П__ЧГ, 1000 ,

^[Zn-^ + S]                        15^10 ³ [/n q +5]

= 0.00056 m3 /s = 48.38 m 3 /day

• 3)    Let's assume that S=20:

_ 2якЛ[(Р _к - P _q ) - i ₀ R _k] _ 2 • 3,14 • 0,1 • 10^-12 • 10[(120 - 80) • 10⁵ - 220 • 1000]

• ^Q        ^[/nR ^ + s]                      15 • 10^-3 [Zn1000 +20]

= 0.00031 m3 /s = 26.78 m 3 /day

Studies show that in most cases, one of the methods used to increase well production is washing the bottom zone with acid. In general, it has been shown in the literature that the amount of acidic solution per 1 meter thickness of the layer is (1.0 ÷ 1.2) m 3 . In this case, the amount of acid:

• V    = sigh

Then we can use the following expression to find the volume of the acidic solution:

^ k

X_p(5,09X_P + 999)

^VX _k (5,09X _k + 999)

Here X is the concentration of the p -acid in the solution

X k -acid (%)

If there is a change in the acid's concentration over time and as the storage conditions change, then the volume of this acid is calculated in the following form:

X_P(5,09X_P + 999)

• V L    = V • 5,09 -       ----^P

P k⁽ P k - ⁹⁹⁹⁾

Here, ρ k is the density of the solution at a temperature of -15 0 C.

P k = P kt + (2,67 • 10^-3p _kt - 2,52)(t - 15)

• ρ kt is the density of the acid at a given temperature.

Barium-chlorine is used to neutralize this acid. Its amount is determined by the following expression:

G_bx = 21,3V (a^- 0,02)

Xk

Here a - indicates the volume of sulfuric acid in the total acid. In mining conditions, this amount is calculated as follows:

a = 0,490

Then we can find the volume of barium-chlorine with the following formula:

Ghx

V, x

P bx

Here p _xb is the density of barium-chlorine, its value is approximately p _xb = 4000 kg/m 3

is accepted. The volume of the inhibitor, in turn, is determined as follows:

• V [ = b [ ^ Here, b i -inhibitor addition rate. If reagent B-2 was used as an inhibitor, then b i =0.2%. c i is the total volume of inhibitor (c i =100%)

Based on this, we calculate the conditional volume of the solution using the following formula:

- х^хз^ ^k            27,5(5,09-27,5+999)

Now let's find the density of the given acid at 15 0C:

P15oC = 1134 + (2,67 - 10-3 - 1134 - 2,52)(25 - 15) = 1139.1 kg/m 3

According to the given temperature, the volume of acid is calculated as follows:

V^ = 12,6 5 • 5,0 9 • 135(5,09125+9222= 5.8 m3

^K                     1139,1(1139,1-999)

Now let's find the amount of barium chloride:

Gbx = 21,3 • 13,2 • (0,4115 — 0,02) = 49,5kg \     2/,D/

The volume of barium chloride is:

14_V = — - 0,0123 « 1,23 • 10^-2m 3

°x

Thus, we get the required volume of technical acid:

V = I ¹²⁶⁵ = 4,74 • 10^-1m 3

Let's calculate the volume of the inhibitor with the help of the formulas mentioned above:

V = 0,2 •¹²⁶⁵ = 2,52 • 10^-2m³

1       ,      100

Volume of the intensifier:

Vin = 0,3 •1265 = 3,8 • 10-2m3 171

Finally, let's calculate the volume of water required:

V s =12.65-5.82-(0.0123+0.474+0.026+0.0379)=6.28 m3